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User:Karlhahn/user pi-irrational

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This user can prove that is irrational.

Usage: {{User:Karlhahn/user pi-irrational}}

PROOF:

If were rational, then

where and are positive integers.

The function, has zeros at and at . So does , where is an arbitrary positive integer -- which is to say that can be chosen to be arbitrarily large. Now we scale the function by to get , which also has zeros at and at . Observe that this function is a th degree polynomial with integer coefficients.

Finally we scale this function by to form

Looking at the derivatives of , we find that the first derivatives also have zeros at and . At the th derivative, Leibniz' rule yields the following:

Observe that all of the coefficients in the last expression are integers. Further observe that

which is an integer, and that

which is also an integer. The same kind of analysis on higher derivatives up to the th derivative shows that they too have all integer coefficients, and more importantly, at and at , they too have integer values. Beyond the th derivative, all derivatives are identically zero. Why? Because is a polynomial of degree .

Now define a new polynomial function, , which is an alternating sum of even derivatives of

It's quite easy to see that . With only a little more effort you can see that

This means that

Since is zero at and at , the right hand side of the above is simply equal to . The previous analysis of derivatives of requires that be an integer. That means that the area under the curve,

between and is an integer. We know that is positive throughout that open interval. We also know that can be bounded in the interval to as small a positive value as you like simply by choosing large enough. The area under the curve can be no greater than times that bound. Hence the area can also be bounded to as small a positive value as you like by choosing large enough. That means the area can be bounded to less than unity. We are left with an area whose value is an integer that is strictly between zero and unity, which is clearly impossible. Hence cannot be rational.